A field extension of prime degree












6












$begingroup$



Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$




Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.



Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$



$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$



Have I inferred it correctly? How do I move ahead?



Thank you for your help.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:44










  • $begingroup$
    $[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:50






  • 1




    $begingroup$
    Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:52










  • $begingroup$
    Ohh I missed seeing that. Thank you for your comment :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55










  • $begingroup$
    @DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
    $endgroup$
    – Qwerty
    Oct 10 '16 at 14:45
















6












$begingroup$



Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$




Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.



Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$



$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$



Have I inferred it correctly? How do I move ahead?



Thank you for your help.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:44










  • $begingroup$
    $[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:50






  • 1




    $begingroup$
    Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:52










  • $begingroup$
    Ohh I missed seeing that. Thank you for your comment :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55










  • $begingroup$
    @DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
    $endgroup$
    – Qwerty
    Oct 10 '16 at 14:45














6












6








6


1



$begingroup$



Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$




Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.



Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$



$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$



Have I inferred it correctly? How do I move ahead?



Thank you for your help.










share|cite|improve this question











$endgroup$





Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~forall~ a in E : ~ F(a)=F$ or $F(a)=E$




Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.



Now, $[E : F(a)] = 1 implies $ there is only one element $x in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = {x~c~|~ c in F(a)}$



$[F(a) : F] = 1 implies $ there is only one element $y in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = {y~d~|~ d in F}$



Have I inferred it correctly? How do I move ahead?



Thank you for your help.







abstract-algebra ring-theory field-theory extension-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 14 '14 at 15:41









user26857

39.4k124183




39.4k124183










asked Aug 14 '14 at 13:39









MathManMathMan

3,66842071




3,66842071








  • 2




    $begingroup$
    The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:44










  • $begingroup$
    $[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:50






  • 1




    $begingroup$
    Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:52










  • $begingroup$
    Ohh I missed seeing that. Thank you for your comment :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55










  • $begingroup$
    @DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
    $endgroup$
    – Qwerty
    Oct 10 '16 at 14:45














  • 2




    $begingroup$
    The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:44










  • $begingroup$
    $[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:50






  • 1




    $begingroup$
    Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
    $endgroup$
    – Daniel Fischer
    Aug 14 '14 at 13:52










  • $begingroup$
    Ohh I missed seeing that. Thank you for your comment :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55










  • $begingroup$
    @DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
    $endgroup$
    – Qwerty
    Oct 10 '16 at 14:45








2




2




$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44




$begingroup$
The thing is that $[K:F] = 1 implies K = F$ for fields $Fsubset K$. Every nonzero element of $K$ gives you a basis, so choose $1$ for simplicity.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:44












$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50




$begingroup$
$[K : F] =1 implies K = {x~c~|~c in F}$ for some $x in K$. Hence, could you explain why every non zero element of $K$ gives a basis?
$endgroup$
– MathMan
Aug 14 '14 at 13:50




1




1




$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52




$begingroup$
Take any $bneq 0$. Then $Fcdot b = { xcdot b : x in F}$ is a one-dimensional subspace of the one-dimensional $F$-vector space $K$, so the full space.
$endgroup$
– Daniel Fischer
Aug 14 '14 at 13:52












$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55




$begingroup$
Ohh I missed seeing that. Thank you for your comment :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55












$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45




$begingroup$
@DanielFischer I am a novice. Can you please explain me what you mean by $Fcdot b$ ?
$endgroup$
– Qwerty
Oct 10 '16 at 14:45










1 Answer
1






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oldest

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2












$begingroup$

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.



Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.



The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.



You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. I get this :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55






  • 1




    $begingroup$
    @VHP Anytime :)
    $endgroup$
    – rschwieb
    Aug 14 '14 at 13:56











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.



Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.



The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.



You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. I get this :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55






  • 1




    $begingroup$
    @VHP Anytime :)
    $endgroup$
    – rschwieb
    Aug 14 '14 at 13:56
















2












$begingroup$

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.



Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.



The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.



You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your answer. I get this :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55






  • 1




    $begingroup$
    @VHP Anytime :)
    $endgroup$
    – rschwieb
    Aug 14 '14 at 13:56














2












2








2





$begingroup$

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.



Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.



The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.



You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$






share|cite|improve this answer









$endgroup$



Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace.



Therefore in $F(a) = {yd mid d in F}$ you may as well choose $y=1in F(a)$ as the generator. Thus $F(a) = {d mid d in F}=F$.



The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this.



You will move ahead by applying this to both $[F(a):F]$ and $[E:F(a)]$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 14 '14 at 13:47









rschwiebrschwieb

107k12102251




107k12102251












  • $begingroup$
    Thank you for your answer. I get this :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55






  • 1




    $begingroup$
    @VHP Anytime :)
    $endgroup$
    – rschwieb
    Aug 14 '14 at 13:56


















  • $begingroup$
    Thank you for your answer. I get this :-)
    $endgroup$
    – MathMan
    Aug 14 '14 at 13:55






  • 1




    $begingroup$
    @VHP Anytime :)
    $endgroup$
    – rschwieb
    Aug 14 '14 at 13:56
















$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55




$begingroup$
Thank you for your answer. I get this :-)
$endgroup$
– MathMan
Aug 14 '14 at 13:55




1




1




$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56




$begingroup$
@VHP Anytime :)
$endgroup$
– rschwieb
Aug 14 '14 at 13:56


















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